# Minimum Length Candidate

Discussion in 'Alternative Theories' started by SimonsCat, Jan 20, 2017.

1. ### SimonsCatRegistered Member

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213

You object to the first equation? (or you object to its presentation as showing any expertise?) How very critical of you.

The first equation is the spherical Schwarzschild metric in the Planck limit. Your objection is unjustified. Likewise, this metric is written for a spacetime uncertainty - which can also be seen as a specific string theory relationship:

$\delta L c\delta t = L^2_P$

If you are curious about this, read up on Yoneya (1987, 1989, 1997), and then when done that, read up on Crowell and how to apply spacetime uncertainty to a metric.

I did my homework, you should go do yours.

3. ### SimonsCatRegistered Member

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I'll try and write a bit more up.

Last edited: Jan 23, 2017

5. ### SimonsCatRegistered Member

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oh, and by the way, that $g_{tt}$ is dimensionless in reply to your study of it.

7. ### SimonsCatRegistered Member

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213
Let's try and retrieve the spherical solution of the metric and hopefully the problem will resolve itself.

$\delta L \delta t\ g_{tt} \geq 1 - \frac{2 G \hbar}{c^4} = \frac{L^2_P}{c}$

Rearranging we have

$\delta L c\delta t\ g_{tt} \geq 1 - \frac{2 G \hbar}{c^3}$

Looking at a metric we have

$ds^2 = (1 - \frac{r_s}{r})^{-1} c^2dt^2$

It seems like the problem is revealing itself.

$\delta L c\delta t\ \geq (1 - \frac{r_s}{r})^{-1}\frac{2 G \hbar}{c^3}$

in which we can define

$g_{tt} = (1 - \frac{r_s}{r})$

(again dimensionless)

So yes, for that reason, it isn't terribly clear at all that first equation, so I suppose you are right. I just never took consideration on it, because it was glanced over by a one of my friends, who mentioned it, but when I said schwarzschild, it was never mentioned again. So it seems, there is an imperfection here, it seems the subtraction of 1 has to drop.

It seems this objections was true.

Last edited: Jan 23, 2017
8. ### SimonsCatRegistered Member

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213
Ok, so I have done the relevant equations above and you were right in your objection of subtraction of 1. Just keep in mind that it is a dimensionless metric.

9. ### rpennerFully WiredStaff Member

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Not that your math is getting any better, but you got this wrong:
when the source you didn't cite, said the metric was:
$ds^2 = \left( 1 - \frac{r_s}{r} \right) c^2 dt^2 \, - \, \left( 1 - \frac{r_s}{r} \right)^{-1} dr^2 \, - \, r^2 d\theta^2 \, - \, r^2 \sin^2 \theta d\phi^2$
which is the famous form of the Schwarzschild metric in normal units using the abbreviation: $r_s \equiv \frac{2 G M}{c^2}$.
Since in geometric units describing coordinates, $ds^2 = g_{tt} (c dt)^2 + g_{tr} (c dt)(dr) + g_{rt} (c dt)(dr) + g_{rr} (dr)^2 + \dots$ we can read off the dimensionless form of $g_{tt}$ as $1 - \frac{r_s}{r}$ which you quote at the bottom.

However you did not use any valid process to go from this:
to this:
. It looks like you are still failing to do the proper thing and start over.

And in reference to that last assertion, so Space-time fluctuations are infinite at the event horizon of a black hole?

10. ### originTrump is the best argument against a democracy.Valued Senior Member

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9,403
So other than that, Reiku is doing his typical bang-up job.

11. ### SimonsCatRegistered Member

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213

I love how critical you are, you are keeping me on my toes. If you want me to do the ''proper thing'' and start over again, I will do so later.

12. ### SimonsCatRegistered Member

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213
Rpenner, this is the correct format. If a moderator could be good enough to move this to the OP with an apology from me.

Spacetime uncertainty is constructed from the metric and it is given as

$\delta L \delta t\ g_{tt} \geq \frac{G \hbar}{c^4} = \frac{L^2_P}{c}$

I showed how this might be suggestive of a relationship between redshift and quantum mechanics via:

$\delta L \delta t \geq \frac{(\frac{G \hbar}{c^4})}{g_{tt}}$

The metric is just

$g_{tt} = 1 + \frac{\phi}{c^2}$

expanding we get

$1 + \phi_1 - \phi_2 - \frac{1}{2} \phi^2_1 - \phi_1\phi_2 + \frac{3}{2}\phi^2_2 +...$

To first order we get

$\delta L \delta t \geq \frac{(\frac{G \hbar}{c^4})}{[\frac{\phi_1}{c^2} - \frac{\phi_2}{c^2}]} = \frac{(\frac{G \hbar}{c^4})}{\frac{\delta \phi}{c^2}}$

adopting $\delta s \equiv c \delta t$ we can theorize the fundamental length as

$\delta L \geq \frac{(\frac{G \hbar}{c^3})}{g_{tt}c\delta t} = \frac{(\frac{G \hbar}{c^3})}{\sqrt{\delta s^2 - g_{xx}\delta x^2}}$

this can be constructed by using

$\delta s^2 = g_{tt} c^2\delta t^2 + g_{xx} dx^2$

and so

$g_{tt}c \delta t = \sqrt{\delta s^2 - g_{xx}\delta x^2}$

Since this formalism is NOT coordinate free, there is clearly work to be done.

13. ### rpennerFully WiredStaff Member

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*WHY* is it constructed from the [Schwarzschild] metric? While the first solution to the GR field equations, it's a pretty crappy metric and isn't even isotropic with respect to the coordinate speed of light. What observation or reasoning do you have to conclude that it is distinct from $\delta L \delta t = \frac{G \hbar}{c^4} = \frac{L^2_P}{c}$ ? Is it a mistake now that the factor of 2 is missing or was it a mistake then? What experimental observation rules out the flat space-time test theory $\delta L \delta t \geq k \frac{G \hbar}{c^4}$ where k=0? where k=10^6? What happens at the black hole event horizon where $g_{tt} = 0$?

Except that doesn't show any such relationship. It's a trivial duplication of the content of the first equation to people who can do algebra.

No, it isn't. It's $g_{tt} = 1 - \frac{r_s}{r} = 1 - \frac{2 GM}{c^2 r} = 1 + \frac{2 \phi(r)}{c^2}$ with $\phi(r) = - \frac{GM}{r}$ only if you ignore that the definition of r in GR differs from the definition of r in Newtonian gravity.

The distance from r=r₁ to r=r₂ at the same angle and time coordinates in the Schwarzschild metric is $\int_{r_1}^{r_2} \sqrt{ g_{rr} } dr = \sqrt{(r_2 - r_s) r_2} - \sqrt{(r_1 - r_s) r_1} + r_s \ln \frac{\sqrt{r_2 - r_s} + \sqrt{r_2} }{ \sqrt{r_1 - r_s} + \sqrt{r_1} } > r_2 - r_1$

$\begin{array}{rr|r|r} r_1 & r_2 & \int_{r_1}^{r_2} \sqrt{ g_{rr} } dr & \left. g_{tt} \right|_{r=r1} \\ \hline \\ 1.000 r_s & 1.100 r_s & 0.643 r_s & 0.000 \\ 1.100 r_s & 1.200 r_s & 0.281 r_s & 0.091 \\ 1.200 r_s & 1.300 r_s & 0.225 r_s & 0.167 \\ 1.300 r_s & 1.400 r_s & 0.197 r_s & 0.231 \\ 1.400 r_s & 1.500 r_s & 0.180 r_s & 0.286 \\ 1.500 r_s & 1.600 r_s & 0.168 r_s & 0.333 \\ 1.600 r_s & 1.700 r_s & 0.159 r_s & 0.375 \\ 1.700 r_s & 1.800 r_s & 0.153 r_s & 0.412 \\ 1.800 r_s & 1.900 r_s & 0.148 r_s & 0.444 \\ 1.900 r_s & 2.000 r_s & 0.143 r_s & 0.474 \\ 2.000 r_s & 3.000 r_s & 1.300 r_s & 0.500 \\ 3.000 r_s & 4.000 r_s & 1.185 r_s & 0.667 \\ 4.000 r_s & 5.000 r_s & 1.135 r_s & 0.750 \end{array}$

So it is oversimplifying (and misleading) to equate $g_{tt} = 1 + \frac{2 \phi(r)}{c^2}$ in the strong gravity regime, where $g_{tt}$ is less than 1 by a appreciable percentage.

Last edited: Jan 24, 2017
14. ### SimonsCatRegistered Member

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213
Yes the 2 should be there, I just got too giddy removing the symbols.

Why do you think gravitational corrections are made in the OP? Did you take my first approximation as a realistic case of quantum gravity? It wasn't meant to be taken as such.

You are asking why it is created from the Schwarzschild metric, can't you actually see why? The structure of the equation will retrieve it.

15. ### SimonsCatRegistered Member

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213
$\delta s^2 \geq (1 - \frac{r_s}{r})^{-1}\frac{2 G \hbar}{c^3}$

This is the schwarschild solution for a spherical body. Clearly from this,

$\delta L c\delta t\ \geq (1 - \frac{r_s}{r})^{-1}\frac{2 G \hbar}{c^3}$

and

$\delta L \delta t\ g_{tt}\geq \frac{2 G \hbar}{c^4}$

hence it is retrievable from that solution.

16. ### rpennerFully WiredStaff Member

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No it isn't. It appears to be the squared length of a small displacement in radial coordinate only where you have chosen a coordinate distance of a Planck length rather than a physical distance of a Planck length. A sphere in such flawed coordinates is a spheroid elongated in the radial direction in physical space.

17. ### SimonsCatRegistered Member

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213
Sorry, can you restate that another way, I am confused by the objection in its totality.

$\delta s^2$ is just a notation for some line element. It doesn't matter that we can later specify: $\delta L c \delta t$ or imply

$\delta L c \delta t = L^2_P$

The right handside, after all, contains exactly the same configuration

$\frac{2G\hbar}{c^3}$

which is of course the Planck length squared anyway, or Planck area. As I stated anyway, the metric was derived very specifically from the work of Crowell, you may want to take a look at this

http://www.ejtp.com/articles/ejtpv9i26p277.pdf